1
$\begingroup$

I find it somewhat unusual that one does not need to define electron coordinates in an electronic structure calculation. Taking Hartree Fock for example, the Fock operator includes the one-electron operator that depends on electron coordinate. However, the matrix representation of the Fock operator in the basis of $\{\phi_{\mu}\}$ is

$F_{\mu v}=\int d \mathbf{r}_{1} \phi_{\mu}^{*}(1) f(1) \phi_{v}(1)$

So because the entries in the Fock matrix are all integrals over all space $\int_{r_{1}=0}^{\infty} d \mathbf{r}_{1}$, it doesn't need the electron coordinate. Is that correct?

Improve this question
$\endgroup$
3
  • $\begingroup$ Electron is not a thing that has coordinates. $\endgroup$
    – Ivan Neretin
    Commented Jan 5, 2022 at 18:16
  • $\begingroup$ Yes, very much so $\endgroup$
    – Ian Bush
    Commented Jan 5, 2022 at 18:16
  • $\begingroup$ @IvanNeretin I'm very much aware of the quantum treatment of electron as a wave rather than a particle with definite coordinates. I'm more concerned with the numerical implementation of electronic structure method! $\endgroup$
    – jayjay
    Commented Jan 5, 2022 at 18:46

2 Answers 2

Reset to default
1
$\begingroup$

The answer is yes and no. Electronic coordinates are needed when calculating the Fock matrix elements $F_{rs}$, which consist of one-electron integrals and two-electron integrals. Once you get the Fock matrix elements, then the Hatree-Fock-Roothaan equations are solved by matrix algebra through diagonalization without using electron coordinates. Therefore, electron coordinates are used explicitly in the step of Fock matrix elements calculations but not in the step of matrix diagonalization.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Please, if you do not think that it does help, i.e. it answers the question, do not post the answer; you may consider a comment instead. If you think your answer is helpful, don't use tag lines like "hope this helps". The voting system is designed to judge the helpfulness of an answer based on reader's opinions. Avoid unnecessary fluff. $\endgroup$
    – Martin - マーチン ♦
    Commented Jan 5, 2022 at 23:31
  • $\begingroup$ It is my understanding that it is humble and polite way to end the answer. Now it is removed. $\endgroup$
    – Josiah_H
    Commented Jan 5, 2022 at 23:38
  • $\begingroup$ I'm sorry to tell you, but it has a negative connotation on the Internet. You might want to look it up on the urban dictionary. $\endgroup$
    – Martin - マーチン ♦
    Commented Jan 5, 2022 at 23:46
  • 1
    $\begingroup$ Point is well taken. I am new here and still learning through mistakes. Thanks. $\endgroup$
    – Josiah_H
    Commented Jan 6, 2022 at 0:04
  • 1
    $\begingroup$ While we're not all new here, we can always still learn a thing or two every day. So don't worry about that. You can also use Chemistry Chat to talk to others, even though it's a bit deserted lately. You can still give it a try, and the system is also right now reminding us to avoid discussions in the comments. So better stop here. Have fun and thanks for your contributions. $\endgroup$
    – Martin - マーチン ♦
    Commented Jan 6, 2022 at 0:10
0
$\begingroup$

Think about the integration as solving the Schrödinger (or Fock, or whatever) equation for all coordinates in space at the same time.

The whole point of this is to obtain a wavefunction which yields the energy eigenvalue to your operator no matter what electron coordinates you plug in. In other terms,

$\hat H \Psi ( \mathbf{x} ) = E \Psi ( \mathbf{x} ) $

for any $\mathbf{x}$.

So, once you have your final wavefunction, nothing stops you from "putting your electrons in specific locations", but that's not really all that interesting because we know that this will still yield the same energy eigenvalue - because that's how we constructed the wavefunction.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.